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I want to know if the first part of my prove (the "if" part) is right. If not, could you show my mistakes please? Let the circle $C$ be given by the equation $$ C: (\mathbf{x}-\mathbf{c})\cdot (\mathbf{x}-\mathbf{c})=r^2,$$ where $\mathbf{c}$ is the center of the circle and $r$ the radius. A line in its parametric form can be given by $$ L= \{\mathbf{u}t + \mathbf{d} |t \in\mathbb{R}\} $$ Therefore, if we choose two points in the line $L$ with values $t$ equal to $s$ and $f$, the polar lines of the points are
$$ S: (\mathbf{u}s + \mathbf{d}- \mathbf{c}) \cdot (\mathbf{x} - \mathbf{c}) = r^2, $$ $$ F: (\mathbf{u}f + \mathbf{d}- \mathbf{c}) \cdot (\mathbf{x} - \mathbf{c}) = r^2 .$$ By substracting the two equations, we obtain $$ (\mathbf{x} - \mathbf{c})\cdot((\mathbf{u}s + \mathbf{d}- \mathbf{c}) - (\mathbf{u}f + \mathbf{d}- \mathbf{c})) = 0$$ $$ (\mathbf{x} - \mathbf{c})\cdot (s-f)\mathbf{u} = (\mathbf{x} - \mathbf{c}) \cdot \mathbf{u} = 0, $$ which is the equation of the line normal to $L$, and passing through the center of the circle $\mathbf{c}$, let this line be $R$. Since the values of $s$ and $f$ are arbitrary, let $s$ fixed and vary $f$, therefore the polar lines of the points of the form $\mathbf{u}f + \mathbf{d}- \mathbf{c}$ must past through the intersection of $S$ and $R$, in other words, they are concurrent.

PD: Definition of a polar line img

The polar line of the point C is the red one

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  • $\begingroup$ You were doing find until the end. Although the intersection of the two polars that you’ve constructed does lie on $R$, it’s not the intersection of $L$ and $R$. Indeed, this point lies in the interior of the circle. $\endgroup$ – amd May 2 '18 at 6:22
  • $\begingroup$ Is it correct now? $\endgroup$ – Pepe Pérez May 2 '18 at 14:43
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Basically correct, but I think there’s a more direct path to the result:

Let $\mathbf p$ and $\mathbf q$ be two distinct points exterior to $C$. Every point $\mathbf r$ colinear with them can be written in the form $(1-\lambda)\mathbf p+\lambda\mathbf q$ for some $\lambda\in\mathbb R$. An equation of the polar line of $\mathbf r$ is $$\left((1-\lambda)\mathbf p+\lambda\mathbf q-\mathbf c\right)\cdot(\mathbf x-\mathbf c)-r^2 = 0.$$ The left-hand side can be rewritten as $$(1-\lambda)[(\mathbf p-\mathbf c)\cdot(\mathbf x-\mathbf c)-r^2]+\lambda[(\mathbf q-\mathbf c)\cdot(\mathbf x-\mathbf c)-r^2],$$ so it is a linear combination of equations for the polars of $\mathbf p$ and $\mathbf q$, therefore passes through their intersection.

If you’re familiar with projective geometry and determinants, this property of pole/polar relationships is pretty much a direct consequence of the definitions. Both colinearity and concurrence can be defined by $\det\small{\begin{bmatrix}a&b&c\end{bmatrix}}=0$. If $C$ is the matrix of a non-degenerate conic, the polar of a point $\mathbf p$ is $C\mathbf p$ and the pole of a line $\mathbf l$ is $C^{-1}\mathbf l$. However, for any matrix $M$, $\det\small{\begin{bmatrix}Ma&Mb&Mc\end{bmatrix}} = \det(M) \det\small{\begin{bmatrix}a&b&c\end{bmatrix}}$, so three points are colinear iff their polar lines are concurrent.

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